HMMT 二月 2019 · 几何 · 第 9 题
HMMT February 2019 — Geometry — Problem 9
题目详情
- In a rectangular box ABCDEF GH with edge lengths AB = AD = 6 and AE = 49, a plane slices through point A and intersects edges BF, F G, GH, HD at points P, Q, R, S respectively. Given that AP = AS and P Q = QR = RS , find the area of pentagon AP QRS .
解析
- In a rectangular box ABCDEF GH with edge lengths AB = AD = 6 and AE = 49, a plane slices through point A and intersects edges BF, F G, GH, HD at points P, Q, R, S respectively. Given that AP = AS and P Q = QR = RS , find the area of pentagon AP QRS . Proposed by: Yuan Yao √ 141 11 Answer: 2 Let AD be the positive x -axis, AB be the positive y -axis, and AE be the positive z -axis, with A the origin. The plane, which passes through the origin, has equation k x + k y = z for some undetermined 1 2 parameters k , k . Because AP = AS and AB = AD , we get P B = SD , so P and S have the same 1 2 z -coordinate. But P (0 , 6 , 6 k ) and S (6 , 0 , 6 k ), so k = k = k for some k . Then Q and R both have 2 1 1 2 49 49 2 2 z -coordinate 49, so Q ( − 6 , 6 , 49) and R (6 , − 6 , 49). The equation QR = RS then gives k k ( ) ( ) 2 2 49 49 2 − 6 + (49 − 6 k ) = 2 12 − − 12 . k k This is equivalent to 2 2 2 (49 − 6 k ) ( k + 1) = 2(49 − 12 k ) , which factors as 3 2 ( k − 7)(36 k − 336 k − 203 k + 343) = 0 . This gives k = 7 as a root. Note that for Q and R to actually lie on F G and GH respectively, we 49 49 must have ≥ k ≥ . Via some estimation, one can show that the cubic factor has no roots in this 6 12 336 28 range (for example, it’s easy to see that when k = 1 and k = = , the cubic is negative, and it 36 3 also remains negative between the two values), so we must have k = 7. Now consider projecting AP QRS onto plane ABCD . The projection is ABCD save for a triangle 49 25 47 ′ ′ Q CR with side length 12 − = 5. Thus the projection has area 36 − = . Since the area k 2 2 of the projection equals [ AP QRS ] · cos θ , where θ is the (smaller) angle between planes AP QRS and ABCD , and since the planes have normal vectors ( k, k, − 1) and (0 , 0 , 1) respectively, we get ( k,k, − 1) · (0 , 0 , 1) 1 1 √ √ √ cos θ = = = and so 2 2 2 k + k +1 2 k +1 99 √ √ 47 99 141 11 [ AP QRS ] = = . 2 2